Vibrations and Waves 2

1 University of Cape Town

Department of Physics

PHY2014F

Vibrations and Waves

Andy Buffler

Department of Physics University of Cape Town

andy.buffler@uct.ac.za

Part 2

Coupled oscillators

Normal modes of continuous systems The wave equation

Fourier analysis

… covering (more or less) French Chapters 2, 5 & 6

Problem-solving and homework

Each week you will be given a take-home problem set to complete and hand in for marks ...

In addition to this, you need to work through the following

problems in French, in you own time, at home. You will not be asked to hand these in for marks. Get help from you friends, the course tutor, lecturer, ... Do not take shortcuts.

Mastering these problems is a fundamental aspect of this course. The problems associated with Part 2 are:

2-2, 2-3, 2-4, 2-5, 2-6, 5-2, 5-8, 5-9, 6-1, 6-2, 6-6, 6-7, 6-10, 6-11, 6-14

3 The superposition of periodic motions

Two superimposed vibrations of equal frequency

1 1 cos( 0 1)

x = A ω t +φ

2 2 cos( 0 2)

x = A ω t +φ

combination can be written as

0

cos( )

x = A ω t +φ

Using complex numbers:

0 1 ( ) 1 1 j t z = A e ω φ+ 0 2 ( ) 2 2 j t z = A e ω φ+ z = +z1 z2

{

}

0 1 2 1 ( ) ( ) 1 2 j t j z e ω φ+ A A e φ φ− ∴ = + A A2 A₁ 0t 1 ω +φ 2 1 φ φ− β 2 1 φ φ φ= − Phase difference Then

(

)

2 2 1 sin sin A β = A φ φ− and 2 2 2 1 2 2 1 2 cos( 2 1) A = A + A + A A φ φ− French page 20

If we add two sinusoids of slightly different frequency and … we observe “beats”… x₁ x₂ x₁+x₂ t t

Superposed vibrations of slightly different frequency: Beats

2 cosω t 1 2 cosω t + cosω t 1 2 cos 2 t ω ω−       2 beat T π ω ω = − 1 cosω t 1 2 1 2 1 2

cos cos 2 cos cos

2 2 t t ω ω t ω ω t ω + ω = _ − _ _ + _     1 ω ω₂ French page 22

5 Combination of two vibrations at right angles

1 cos( 1 1)

x = A ω t +φ

2 cos( 2 2)

y = A ω t +φ ???

Write andx = A₁ cos(ω₀t)

Consider case where frequencies are equal and let initial phase difference be 2 cos( 0 ) y = A ω t +φ Case 1 : φ = 0 x = A₁ cos(ω₀t) 2 cos( 0 ) y = A ω t 2 1 A y x A = Rectilinear motion 2 φ π= x = A₁ cos(ω₀t) 2 cos( 0 2) 2 sin( 0 ) y = A ω t +π = −A ω t Case 2 : 2 2 2 2 1 2 1 x y A A ∴ + = Elliptical path in clockwise direction French page 29 φ

Case 3 : φ π= x = A₁ cos(ω₀t) ₂ 1 A y x A = −

Combination of two vibrations at right angles …2

2 cos( 0 ) 2 cos( 0 ) y = A ω t +π = −A ω t 3 2 φ = π x = A₁ cos(ω₀t) 2 cos( 0 3 2) 2 sin( 0 ) y = A ω t + π = +A ω t Case 4 : 2 2 2 2 1 2 1 x y A A ∴ + = − Elliptical path in anticlockwise direction 4 φ π= x = A₁ cos(ω₀t) 2 cos( 0 4) y = A ω t +π Case 5 :

7

Superposition of simple harmonic vibrations at

right angles with an initial phase difference of π 4

Superposition of two perpendicular simple harmonic motions of the same frequency for

9

Abbreviated construction for the superposition of vibrations at right angles … see French page 34.

Perpendicular motions with different frequencies: Lissajous figures

See French page 35.

Lissajous figures for with various initial phase differences.

2 2 1

11 2 : 1 ω ω 1:1 1:2 2:3 1:3 3:4 5:6 4:5 3:5 φ = 0 π 4 π 2 3π 4 π Lissajous figures

Coupled oscillators

When we observe two weakly coupled identical oscillators A and B, we see:

t

t x_A

x_B

… these functions arise mathematically from the addition of two SHMs of similar frequencies … so what are these two SHMs?

These two modes are known as normal modes … which are states of the system in which all parts of the system oscillate with SHM

French page 121

13 t t A B x_A x_B Coupled oscillators

The double mass-spring oscillator m A x k m B x k

Individually we know that and mx_A = −kx_A mx_B = −kx_B For both oscillators: ₀ k

m

ω =

Now add a weak coupling force:

m A x k m B x k c k For mass A: mx_A = −kx_A + k x_c( _B − x_A) 2 2 ω = − + Ω −  ω2 = k Ω =2 k_c

15 2 2 0 ( ) A A B A x = −ω x + Ω x − x 

The double mass-spring oscillator …2

For mass A:

For mass B: xB = −ω02xB − Ω2(xB − xA)

… two coupled differential equations … how to solve ? Adding them: 2 2 0 2 ( A B) ( A B) d x x x x dt + = −ω + Subtract B from A: 2 2 2 0 2 ( A B) ( A B) 2 ( A B) d x x x x x x dt − = −ω − − Ω −

Define two new variables:

1 A B q = x + x 2 A B q = x − x Then and 2 2 1 0 1 2 d q q dt = −ω 2 2 2 2 0 2 2 ( 2 ) d q q dt = − ω + Ω … called “normal coordinates”

The double mass-spring oscillator …3

The two equations are now decoupled …

2 2 1 1 2 s d q q dt = −ω Write ₂ 2 2 2 2 f d q q dt = −ω 2 2 0 s ω = ω 2 2 2 0 2 f ω = ω + Ω s = “slow” f = “fast”

… which have the solutions:

1 cos( s 1)

q = C ω t +φ

2 cos( f 2)

q = D ω t +φ

Since and q₂ = x_A − x_B We can write and

1 A B q = x + x

(

)

1 1 2 2 A x = q + q 1

(

)

1 2 2 B x = q − q

17 The double mass-spring oscillator …4

1 2 cos( ) cos( ) 2 2 A s f C D x = ω t +φ + ω t +φ Then 1 2 cos( ) cos( ) 2 2 B s f C D x = ω t +φ − ω t +φ

So x_A and x_B have been expressed as the sum and difference of two SHMs as expected from observation.

… C, D, φ₁ and φ₂ may be determined from the initial conditions. … when x_A = x_B ,then q₂ = 0 … there is no contribution from the

fast mode and the two masses move in phase … the coupling spring does not change length and has no effect on the motion … ω_s = ω₀

… when x_A = −x_B ,then q₁ = 0 … there is no contribution from the slow mode … the coupling spring gives an extra force … each mass experiences a force giving − +

(

k 2k_c

)

x

2 2 0 2 ω = + Ω 2 2 _c f k k m ω = +

The double mass-spring oscillator …5

19 The double mass-spring oscillator …6

We now have a system with two natural frequencies, and experimentally find two resonances.

Frequency A m pl it ude

Pitch and bounce oscillator d A x k m B x L k

Two normal modes (by inspection): Bouncing A B x = x Restoring force = −2kx 2 2 2 d x m kx dt ∴ = − Pitching A B x = −x xA B x θ

Centre of mass stationary

2 bounce 2k m ω = I τ = − θ 2 1θ = − 1 θ 2 6kd θ θ → = ω2 = 6k d2 French page 127

21 N = 2

(

)

1 0 0 1 2 sin 2 2 1 π ω = ω = ω +

(

)

2 0 0 2 2 sin 3 2 2 1 π ω = ω = ω +

N = 4 N = 3

23 N-coupled oscillators fixed fixed Tension T l 1 2 3 p−1 p p+1 N … consider transverse displacements that are small.

Each bead has mass m

1 2 3 p−1 p p+1 N

1

p

α ₋ αp

Transverse force on pth particle: F_p = −T sinα_p−1 +T sinα_p

1 1 p p p p y y y y T T l l − + − − = − + for small α y French page 136

N-coupled oscillators …2 2 1 1 2 p p p p p p d y y y y y F m T T dt l l − + − − = = − +

(

)

2 2 2 0 0 1 1 2 2 0 p p p p d y y y y dt ω ω + − ∴ + − − = where ,₀2 T ml ω = p = 1, 2 … N

… a set of N coupled differential equations. Normal mode solutions: y_p = A_p sinωt

Substitute to obtain N simultaneous equations

(

2 2

)

2

(

)

0 0 1 1 2 A_p A_p A_p 0 ω ω ω _{+ −} − + − + = or 2 2 1 1 2 ₀ p p A ₊ + A ₋ −ω + ω =

25

N-coupled oscillators …3

From observation of physical systems we expect sinusoidal “shape functions” of the form A_p = Csin pθ

Substitute into 2 2 1 1 ₀ 2 0 2 p p p A A A ω ω ω + + − ₌ − +

And apply boundary conditions and_A0 = ₀ A_N+1 = 0

and n = 1, 2, 3, … N (modes)

(

)

0 2 sin 2 1 n n N π ω = ω + … find that 1 n N π θ = + There are N modes:

sin sin sin

1 pn pn n n n pn y A t C t N π ω ω = = +

0 2ω n ω 0 1 2 3 N+1 n

(

)

0 2 sin 2 1 n n N π ω = ω + N-coupled oscillators …4 For small N:

27 For n << N :

(

)

(

)

sin 2 1 2 1 n n N N π π = + +

(

)

0 0 2 2 1 1 n n n N N πω π ω = ω _{= } _ + _ + _ then i.e. ω_n ∝ n for n << N In many systems of interest N is very large… and we are only interested in the lowest frequency modes. 0 2ω n ω 0 n << N N+1 n linear region N-coupled oscillators …5

N coupled oscillators have N normal modes and hence N resonances

response

ω

29 Continuous systems

Continuous systems

Consider a string stretched between two rigid supports …

x = 0 _{x = L}

tension T

String has mass m and mass per unit length µ = m L Suppose that the string is disturbed in some way:

y x

31 T T θ θ + ∆θ y x _x + ∆_{x x}

Consider the forces on a small length of string …

Restrict to small amplitude disturbances … then is small andθ

cosθ =1 sin tan y

x

θ = θ θ= = ∂ ∂ The tension T is uniform throughout the string.

Net horizontal force is zero: cos(T θ + ∆ −θ) T cosθ = 0 Vertical force: _F = _{T sin(}θ + ∆ −θ_{) T sin}θ

Then F T y _{x x} T y _x

x +∆ x

∂ ∂

= −

∂ ∂

Normal modes of a stretched string

French page 162

x x x y y F T T x +∆ x ∂ ∂ = − ∂ ∂ Use dg g x( x) g x( ) dx x + ∆ − = ∆ Then 2 2 y F T x x ∂ = ∆ ∂ or

(

)

2 2 2 2 y y x T x t x µ∆ ∂ = ∂ ∆ ∂ ∂ giving 2 2 2 2 y y x T t µ ∂ _{ } ∂ =   ∂ _{ } ∂

Check: has the dimensions T µ 2 1 v Then is the speed at which a wave propagates along the string … see later

v = T µ Write 2 2 2 2 2 1 y y x v t ∂ ∂ = ∂ ∂

Normal modes of a stretched string …2

33

Look the standing wave (normal mode) solutions …

Normal mode: all parts of the system move in SHM at the same frequency …

Write: ( , )y x t = f x( ) cosωt ( )

f x is the “shape” function … substitute into wave equation

2 2 2 2 ( , ) ( ) cos y x t d f x t x dx ω ∂ = ∂

(

)

2 2 2 ( , ) ( ) cos y x t f x t t ω ω ∂ = − ∂ 2 2 2 2 ( ) 1 cos ( ) cos d f x t f x t dx ω = − v ω ω

which must be true for all t

then 2 2 2 2 ( ) d f f x dx v ω = −

2 2 2 2 ( ) d f f x dx v ω = −

… which has the same form as the eq. of SHM:

2 2 0 2 d x x dt = −ω … has general solution: _x = _Asin(ω_0t +φ₎

Thus we must have: f x( ) Asin x v

ω

 

= _ + Φ_

 

Apply boundary conditions: y = 0 at x = 0 and x = L

(0) 0 f ∴ = and f L( ) = 0 x = 0, f =0 : 0 Asin 0 v ω   = _ + Φ_   i.e. 0 = Asin_ω L_ 0 Φ = x = L, f =0 : i.e. ω L = nπ n = 1,2,3,…

35 n n v L π ω = Write n = 1,2,3,… ( ) _n sin x n v _n sin n x f x A A v L L π π     = _{ } = _{ }     Therefore

shape function, or “eigenfunction”

x = 0 x = L n = 1 n = 3 n = 2 n = 5 n = 4

(

)

1 ( ) sin f x = A π x L

(

)

2 ( ) sin 2 f x = A π x L

(

)

3 ( ) sin 3 f x = A π x L

(

)

4 ( ) sin 4 f x = A π x L

(

)

5 ( ) sin 5 f x = A π x L 1 v L ω π= 3 3 v L ω = π 2 2 v L ω = π 5 5 v L ω = π 4 4 v L ω = π

Normal modes of a stretched string

n = 1

n = 2

n = 3

Full solution for our standing waves: ( , ) _n sin n x cos _n y x t A t L π _ω   = _{ }   n n v L π ω =

The mode frequencies are evenly spaced: ω_n = nω₁

n ω 3 ω 2 ω 1 ω

(recall the beaded string)

This continuum approach breaks down as the wavelength approaches atomic dimensions… also if there is any “stiffness” in the spring which adds an additional restoring force which is more pronounced in the high frequency modes.

39

All motions of the system can be made up from the superposition of normal modes

1 ( , ) _n sin cos( _{n n}) n n x y x t A t L π _{ω φ} ∞ =   = _{ } +  

∑

with n n v L π ω =

Note that the phase angle is back since the modes may not be in phase with each other.

Whispering galleries

… best example is the inside dome of St. Paul’s cathedral. If you whisper just inside the dome, then an observer close to you can hear the whisper

coming from the opposite direction … it has travelled right round the inside of the dome.

41 Longitudinal vibrations of a rod

x x + ∆x x + + ∆ + ∆ξ x ξ x +ξ 1 F 2 F

section of massive rod

section is displaced and stretched by an unbalanced force Average stress = x ξ ∆ ∆ Average strain = _Y x ξ ∆ ∆ Y : Young’s modulus

stress at = (stress at x) + x + ∆x (stress) x x

∂ _∆

∂

French page 170

If the cross sectional area of the rod is α 1 F Y x ξ α ∂ = ∂ 2 2 ₂ F Y Y x x x ξ ξ α ∂ α ∂ = + ∆ ∂ ∂ and 2 1 2 ₂ F F Y x ma x ξ α ∂ − = ∆ = ∂ 2 2 2 2 Y x x x t ξ ξ α ∂ ρα ∂ ∴ ∆ = ∆ ∂ ∂ or 2 2 2 2 x Y t ξ ρ ξ ∂ ∂ = ∂ ∂ 2 2 2 2 2 1 x v t ξ ξ ∂ ∂ ∴ = ∂ ∂ Y v ρ =

43

Apply boundary conditions: one end fixed and the other free

x = L :

x = 0 : i.e. Φ = 0

n = 1,2,3,… ( , )x t f x( ) cos t

ξ = ω

Look for solutions of the type:

(0, )t 0 ξ = 0 F Y x ξ α ∂ = = ∂ ( ) sin f x A x v ω   = _ + Φ_   where … then cos L 0 v ω  _{ =}     or L

(

n 1₂

)

v ω _π = −

(

1

)

(

1

)

2 2 n n v n _Y L L π π ω ρ − − = =

The natural angular frequencies

x = 0 x = L n = 1 n = 3 n = 2 n = 5 n = 4 1 2 Y L π ω ρ = 2 3 2 Y L π ω ρ = 3 5 2 Y L π ω ρ =

45

n = 1 n = 2 n = 3

Normal modes for different boundary conditions

Simply supported Clamped one end Free both ends Clamped both ends

The elasticity of a gas l A ρ, p Bulk modulus: dp K V dV = −

Kinetic theory of gases: Pressure 13 rms2 rms2

3 m p v v Al ρ = = If thenE_k = 1₂ mv_rms2 2 3 k E p A l =

Now move piston so as to compress the gas … work done on gas:

k W pA l E ∆ = − ∆ = ∆ Then

(

)

2 2 2 2 5 ( ) 3 3 3 3 k k E l l l p E pA l p p A l A l A l l ∆ ∆ ∆ ∆ ∆ = − ∆ = − ∆ − = − adiabatic 5 3 p K V p V ∆ = − = ∆ giving and v = K = 1.667 p French page 176

47 Sound waves in pipes

A sound wave consists of a series of compressions and rarefactions of the supporting medium (gas, liquid, solid) In this wave individual molecules move longitudinally with

SHM. Thus a pressure maximum represents regions in which the molecules have approached from both sides, receding from the pressure minima.

wave propagation

French page 174

49 Pressure p p₀ Flow velocity u 0 x x x x 0 : t = 2 : t = T p u

Standing sound waves in pipes …2

Consider a sound wave in a pipe. At the closed end the flow velocity is zero (velocity node, pressure antinode).

At the open end the gas is in contact with the atmosphere, i.e. p = p₀ (pressure node and velocity antinode).

p

u p₀

0

pressure node _{pressure antinode}

51 Pipe closed at both ends Pipe open at both ends Pipe open at one end n n v L π ω = 2 2 n nv L f λ = = 2 nv f L =

(

2 1

)

(

2 1

)

4 4 n n _v L f λ − − = =

(

2 1

)

4 n v f L − =

(

2 1

)

2 n n v L π ω = −

Sound

Audible sound is usually a longitudinal compression wave in air to which the eardrum responds.

Velocity of sound (at NTP) ~ 330 m s-1

By considering the transport of energy by a compression wave,

can show that 2 2 2

2 _m

P = π f ρ Avs

… where A is cross sectional area of air column and s_m is maximum displacement of air particle in longitudinal wave Then intensity = P ₂ 2 _f 2 _vsm2

A = π ρ unit: W m

53

The human ear detects sound from ~10-12 W m-2 to ~1 W m-2 … use a logarithmic scale for I :

10 0 10 log I I β =  _{ }   decibels

where I₀ = “reference intensity” = 10-12 W m-2 Intensity level or “loudness”:

Musical sounds

Waveforms from real musical instruments are complex, and may contain multiple harmonics, different phases, vibrato, ...

Pitch is the characteristic of a sound which allows sounds to be ordered on a scale from high to low (!?). For a pure tone, pitch is determined mainly by the frequency, although sound level may also change the pitch. Pitch is a subjective sensation and is a subject in “psychoacoustics”.

The basic unit in most musical scales is the octave. Notes judged an octave apart have frequencies nearly (not exactly) in the ratio 2:1. Western music normally divides the octave into 12 intervals called semitones ... which are given note names (A through G with sharps and flats) and designated on musical scales.

55

Timbre is used to denote “tone quality” or “tone colour” of a sound and may be understood as that attribute of auditory

sensation whereby a listener can judge that two sounds are dissimilar using any criteria other than pitch, loudness or duration. Timbre depends primarily on the spectrum of the stimulus, but also on the waveform, sound pressure and temporal characteristics of the stimulus.

Musical sounds ...2

One subjective rating scale for timbre (von Bismarck, 1974) dull compact full colourful sharp scattered empty colourless

Two dimensional systems

… the membrane has mass per unit area , and a surface tension S which gives a force SΔl perpendicular to a length Δl in the surface …

y

x Δy

Δx

The forces on the shaded portion are … SΔx SΔy SΔy Consider an elastic membrane clamped at its edges … French page 181 σ

57

If the membrane is displaced from the z = 0 plane then a cross section through the shaded area shows:

θ θ + ∆θ z x _x + ∆_{x x} SΔy SΔy

… looks exactly like the case of the stretched string.

2 2 z S y x x ∂ ∆ ∆ ∂ The transverse force on the element will be

And if we looked at a cross section perpendicular to this …

the transverse force will be 2

2 z S x y y ∂ ∆ ∆ ∂

The mass of the element is .σ∆ ∆x y Thus 2 2 2 2 2 2 z z z S y x S x y x y x y σ t ∂ ∂ ∂ ∆ ∆ + ∆ ∆ = ∆ ∆ ∂ ∂ ∂ or 2 2 2 2 2 2 z z z x y S t σ ∂ ∂ _{ } ∂ + _{=  } ∂ ∂ _{ } ∂

… a two dimensional wave equation

… with the wave velocity _v S

σ

=

59 ( , , ) ( ) ( ) cos z x y t = f x g y ωt 2 2 2 2 ( ) cos z d f g y t x dx ω ∂ ₌ ∂

(

)

2 2 2 ( ) ( ) cos z f x g y t t ω ω ∂ = − ∂

Look for normal mode solutions of the form:

2 2 2 2 ( ) cos z d g f x t y dy ω ∂ ₌ ∂ 2 2 2 ( ) cos 2 ( ) cos d f d g g y t f x t dx ω + dy ω = 2 2 f x g y( ) ( ) cos t v ω _ω − 2 2 2 2 2 2 1 d f 1 d g f dx g dy v ω + = − i.e.

In a similar fashion to the 1D case, find …

1 1 ( ) _n sin x n x f x A L π   = _{ }   and 2 2 ( ) _n sin y n y g y B L π   = _{ }  

…then

1 2 1 2

1 2

,

( , , ) _{n n} sin sin cos _{n n}

x y n x n y z x y t C t L L π  π  _ω   = _{   }   _{ } 1 2 2 2 1 2 , n n x y n v n v L L π π ω = _ _{ + }_ _   _{ }

where the normal mode frequencies are

e.g. for a membrane having sides 1.05L and 0.95L then 1 2 2 2 1 2 , 1.05 0.95 n n n n v L π ω = _ _ + _{ }    

61 up down 1,1 2,1 2,2 3,1 3,2

Normal modes of a circular membrane

63

Modes of vibration of a 38 cm cymbal. The first 6 modes resemble those of a flat plate ... but after that the resonances tend to be combinations of two or more modes.

65 Chladni plates

67

Holographic interferograms of the top and bottom plates of a violin at several resonances.

Holographic interferograms of a classical guitar top plate at several resonances.

69

Holographic interferograms showing the vibrations of a 0.3 mm thick trombone driven acoustically at 240 and 630 Hz.

71

one point per normal mode 1 2 2 2 1 2 , 2 2 n n n v n v f L L     = _{ } +_{ }    

Normal modes of a square membrane

1 2 v L 2 2 v L 3 2 v L 4 2 v L 0 1 2 v L 2 2 v L 3 2 v L 4 2 v L 5 2 v L 0 0 1 2 3 4 5 0 1 2 3 4 4,3 f 1 n 2 n

Normal modes having the same frequency are said to be degenerate

area per point =

2 2 v L      

Normal modes of a square membrane … for large n₁ and n₂ 1 n 2 n df f area = 2 2 v L      

area per mode =

1

4 (2π f df)

Number of modes with

frequencies between f and (f + df) =

2 1 4 2 (2 f df) L v π _ _   2 2π L f df =

73 Three dimensional systems

Consider some quantity Ψ which depends on x , y , z and t , e.g. the density of air in a room.

In three dimensions: 2 2 2 2 2 2 2 2 2 1 x y z v t ∂ Ψ ∂ Ψ ∂ Ψ_{+ + =} ∂ Ψ ∂ ∂ ∂ ∂

which can be written:

2 2 2 2 1 v t ∂ Ψ ∇ Ψ = ∂ The solutions for a rectangular enclosure:

1 2 3 2 2 2 3 1 2 , , n n n x y z n v n v n v L L L π π π ω = _ _ + _ _ +_ _     _{ }

…and for a cube:

1 2 3 2 2 2 , , 1 2 3 n n n v n n n L π ω = + + French page 188

How many modes are there with frequencies in the range f and (f + df) … ?

Set up an imaginary cubic lattice with spacing v 2L

2 n 1 n 3 n df f

… and consider positive frequencies only.

Volume of shell = 1₈ (4π f 2)df 3 2 v L      

Volume per mode =

Number of modes with

frequencies between f and (f + df) =

3 2 1 8 2 (4 f )df L v π _ _   3 2 4π L f df =

75

Number of modes with

frequencies between f and (f + df)

2 3 4 V f df v π =

… holds for any volume V … provided its dimensions are much greater than the wavelengths involved.

… need to multiply by a factor of 2 when dealing with electromagnetic radiation (2 polarization states) …

“Ultraviolet catastrophe” for blackbody radiation …

Equipartition theorem: in thermal equilibrium each mode has an average energy in each of its two energy stores

Hence, energy density of radiation in a cavity:

1 2 k TB

(

)

2 1 2 3 4 2 V f df 2 df kT c π µ = or 2 3 8 f kT c π µ = f µ experiment_!?

Planck was able to show, effectively by assuming that energy was emitted an

absorbed in quanta of energy hf , that the average energy of a cavity mode was not kT but

1

hf kT

hf

e −

where Planck’s constant h = 6.67 10-34 J K-1 Then 2 3 8 1 hf kT f df hf df c e π µ = − energy density no. of modes in range f to f +df average energy per mode Planck’s law

Introduction to Fourier methods

We return to our claim that any physically observed shape function of a stretched string can be made up from normal mode shape functions.

x ( ) f x i.e. 1 ( ) _n sin n n x f x B L π ∞ =   = _{ }  

∑

… a surprising claim … ? … first find B_n … multiply both sides by

and integrate over the range x = 0 to x = L

1

sin

n x

L

π













1 1

( )sin sin sin

L L n x n x n x f x π dx π B π dx ∞   ₌          

∑

∫

∫

French page 189

79

1 1

1

0 0

( )sin sin sin

L L n n n x n x n x f x dx B dx L L L π π ∞ π =   ₌           _{ }    

∑

∫

∫

Fourier methods …2

If the functions are well behaved, then we can re-order things:

1 1

1

0 0

( )sin sin sin

L L n n n x n x n x f x dx B dx L L L π ∞ π π =   ₌           _{ }  

∑

 

∫

∫

[n₁ is a particular integer and n can have any value between 1 and .] ∞ Integral on rhs:

(

1

)

(

1

)

1

0 0

1

sin sin cos cos

2 L L n n x n n x n x n x dx dx L L L L π π π π _  −  + _     _{= −}             _{   }

∫

∫

Fourier methods …3

Both (n₁ + n) and (n₁ − n) are integers, so the functions

on the interval x = 0 to L must look like

… from which it is evident that

(

1

)

cos n n x L π  ±     

(

1

)

0 cos 0 L n n x dx L π  ±  =    

∫

Except for the special case when n₁ and n are equal … then …

(

1

)

cos n n x 1 L π  −  =     and

(

1

)

cos L n n x dx L L π  −  =    

∫

81 Fourier methods …4

Thus all the terms in the summation are zero, except for the single case when n₁ = n _i.e.

1 2 n L B =

(

)

(

)

1 1 1 1 0 0 1

( )sin cos cos

2 L L n n n x n n x n x f x dx B dx L L L π π π _  −  + _   _{= −}         _{   }

∫

∫

1 1 0 2 ( )sin L n n x B f x dx L L π   = _{ }  

∫

i.e.

We have found the value of the coefficient for some

particular value of n₁ … the same recipe must work for any value, so we can write:

0 2 ( )sin L n n x B f x dx L L π   = _{ }  

∫

Fourier methods …5

The important property we have used is that the functions and

are “orthogonal over the interval x = 0 to x = L.”

i.e. 1 sin n x L π       sin n x L π       1 0 sin sin L n x n x dx L L π π     ₌        

∫

0 if n1 ≠ n 1 if 2 L n = n

83 Fourier methods …6

The most general case (where there can be nodal or antinodal boundary conditions at x = 0 and x = L) is

0 1 ( ) cos sin 2 _n n n A n x n x f x A B L L π π ∞ =  _{   } = + _{  } + _{ }      

∑

where 0 2 ( )sin L n n x B f x dx L L π   = _{ }  

∫

0 2 ( ) cos L n n x A f x dx L L π   = _{ }  

∫

Fourier methods …7

One of the most commonly encountered uses of Fourier methods is the representation of periodic functions of time in terms of sine

and cosine functions …

Put 2

T

π

Ω =

This is the lowest frequency in … clearly there are

higher frequencies … by the same method as before, write … ( ) f t 0 1 2 2 ( ) cos sin 2 _n n n A nt nt f t A B T T π π ∞ =     = + _{ } + _{ }    

∑

0 1 cos sin 2 _n n n A A n t B n t ∞ = = +

∑

Ω + Ω

where and 2 ( )sin

(

)

T n B =

∫

f t n t dtΩ

(

)

2 ( ) cos T n A =

∫

f t n t dtΩ T ( ) f t t

85 Waveforms of ... a flute a clarinet an oboe a saxophone

87

89

91 Odd functions

An odd periodic function _{f (}− = −_{t) f t( )} where −T 2 < <t T 2

t t t

( )

f t f t( ) f t( )

…can be expressed as a sum of sine functions

0 ( ) _n sin n f t B n t ∞ = =

∑

Ω 2 T π Ω =

(

)

0 2 ( )sin T n B f t n t dt T =

∫

Ω

Even functions

An even periodic function f (− = +t) f t( ) where −T 2 < <t T 2

t t t

( )

f t f t( ) f t( )

…can be expressed as a sum of cosine functions

2 T π Ω = 0 1 ( ) cos 2 _n n A f t A n t ∞ = = +

∑

Ω

(

)

2 ( ) cos T A =

∫

f t n t dtΩ

93 Fourier methods … Example

Find Fourier coefficients for the case: ( )

f t

t 1

-1

This is an odd function: ( )f − = −t f t( )

1 ( ) _n sin n f t B n t ∞ = ∴ =

∑

Ω

(

)

0 2 ( )sin T n B f t n t dt T =

∫

Ω

(

)

(

)

2 0 2 2 2 (1)sin ( 1)sin T T T n t dt n t dt T T =

∫

Ω +

∫

− Ω

(

)

2

(

)

0 2 2 1 2 1 (1) cos T ( 1) cos T T n t n t T n T n  _{ }  _{ } = _− _ Ω _ + − _− _ Ω _ Ω Ω    

Fourier methods … Example cont. 2 T π Ω =

[

]

[

]

1 1

cos cos 0 cos 2 cos

n

B n n n

nπ π nπ π π

∴ = − − + −

For even n: cos 2nπ = cosnπ =1 ∴B_{even n} = 0 For odd n: and cosnπ = −1 cos 2nπ = +1

odd 1 1 4 ( 1 1) (1 1) n B nπ nπ nπ ∴ = − − − + + = 4 1 1

( ) sin sin 3 sin 5 ...

3 5 f t t t t π   ∴ = _ Ω + Ω + Ω + _  

95 Fourier sums … Example 3

2 terms 8 terms 4 terms 200 terms 50 terms 20 terms 2 T 2 T −

Fourier sums … Example 1 T 2 terms 4 terms 3 terms 50 terms 20 terms 8 terms 0

97 Fourier sums … Example 2

T 2 terms 8 terms 4 terms 200 terms 50 terms 20 terms 0

t t t t t t f f f f f f

Time domain Frequency spectrum

1 f 1 3 5 71 1 1 f f f f 1 7 f Fourier transforms